Cantor Set and Its Properties - UC Santa Barbaraweb.math.ucsb.edu/~padraic/ucsb_2013_14/mathcs103_s2014/mathcs103... · TitleAbstractPreliminariesConstruction and FormulaProperties

Title Abstract Preliminaries Construction and Formula Properties and Proofs

Cantor Set and Its Properties

Zhixing Guo

University of California, Santa Barbara

April 23, 2014


Abstract

The Cantor set is a famous set first introduced by Germanmathematician Georg Cantor in 1883. It is simply a subset ofthe interval [0,1], but it has a number of remarkable and deepproperties. We will first describe the construction and theformula of the Cantor ternary set, which is the most commonmodern construction, and then prove some interestingproperties of the set.


Preliminaries

Definition: If A and B are sets, the union of A and B, writtenA ∪ B, is the set of all objects that belong to either A or B orboth. A ∪ B = {x : x ∈ A or x ∈ B}

Definition: If A and B are sets, the intersection of A and B,written A ∩ B, is the set of all objects that belong to both A andB. A ∩ B = {x : x ∈ A and x ∈ B}

Example:Let A = {1,2,3},B = {2,3,4}. Then

A ∩ B = {2,3},A ∪ B = {1,2,3,4}


Preliminaries

Definition: If A and B are sets, the union of A and B, writtenA ∪ B, is the set of all objects that belong to either A or B orboth. A ∪ B = {x : x ∈ A or x ∈ B}

Definition: If A and B are sets, the intersection of A and B,written A ∩ B, is the set of all objects that belong to both A andB. A ∩ B = {x : x ∈ A and x ∈ B}

Example:Let A = {1,2,3},B = {2,3,4}. Then

A ∩ B = {2,3},A ∪ B = {1,2,3,4}


Definition: Let Λ be a set, ans suppose for each λ ∈ Λ, asubset Aλ of a given set S is specified. The collection of setsAλ is called an indexed family of subsets of S with Λ as theindex set. We denote this by {Aλ}λ∈Λ

Definition: If {Aλ}λ∈Λ is an indexed family of sets, define⋂λ∈Λ

Aλ = {x : x ∈ Aλfor allλ ∈ Λ}

⋃λ∈Λ

Aλ = {x : x ∈ Aλfor someλ ∈ Λ}

Example:Let An = {n} for each n ∈ N. Then

∞⋂n=1

An = ∅

∞⋃n=1

An = N


Definition: Let Λ be a set, ans suppose for each λ ∈ Λ, asubset Aλ of a given set S is specified. The collection of setsAλ is called an indexed family of subsets of S with Λ as theindex set. We denote this by {Aλ}λ∈Λ

Definition: If {Aλ}λ∈Λ is an indexed family of sets, define⋂λ∈Λ

Aλ = {x : x ∈ Aλfor allλ ∈ Λ}

⋃λ∈Λ

Aλ = {x : x ∈ Aλfor someλ ∈ Λ}

Example:Let An = {n} for each n ∈ N. Then

∞⋂n=1

An = ∅

∞⋃n=1

An = N


Construction

The Cantor ternary set is created by repeatedly deleting theopen middle thirds of a set of line segments.

One starts by deleting the open middle third(

13,23

)from the

interval [0,1], leaving two line segments:[0,

13

]∪[

23,1].

Next, the open middle third of each of these remainingsegments is deleted, leaving four line segments:[0,

19

]∪[

29,13

]∪[

23,79

]∪[

89,1].

This process is continued to infinity.


Construction



13,23

)from the


13

]∪[

23,1].


19

]∪[

29,13

]∪[

23,79

]∪[

89,1].



Construction



13,23

)from the


13

]∪[

23,1].


19

]∪[

29,13

]∪[

23,79

]∪[

89,1].



Construction



13,23

)from the


13

]∪[

23,1].


19

]∪[

29,13

]∪[

23,79

]∪[

89,1].



The picture below shows the firs four steps of this process:


The picture below shows the firs four steps of this process:


FormulaConsider the following set:

C =∞⋂

n=1

3n−1−1⋂k=0

([0,

3k + 13n

]∪[

3k + 23n ,1

])

Does Cantor ternary set contain set C? Yes.

Does set C contain Cantor ternary set? Yes.

Thus, set C is Cantor ternary set.

Another explicit formula for Cantor set is

C = [0,1] \∞⋃

n=1

3n−1−1⋃k=0

(3k + 1

3n ,3k + 2

3n

)



C =∞⋂

n=1

3n−1−1⋂k=0

([0,

3k + 13n

]∪[

3k + 23n ,1

])Does Cantor ternary set contain set C?

Yes.




C = [0,1] \∞⋃

n=1

3n−1−1⋃k=0

(3k + 1

3n ,3k + 2

3n

)



C =∞⋂

n=1

3n−1−1⋂k=0

([0,

3k + 13n

]∪[

3k + 23n ,1

])Does Cantor ternary set contain set C? Yes.




C = [0,1] \∞⋃

n=1

3n−1−1⋃k=0

(3k + 1

3n ,3k + 2

3n

)



C =∞⋂

n=1

3n−1−1⋂k=0

([0,

3k + 13n

]∪[

3k + 23n ,1


Does set C contain Cantor ternary set?

Yes.



C = [0,1] \∞⋃

n=1

3n−1−1⋃k=0

(3k + 1

3n ,3k + 2

3n

)



C =∞⋂

n=1

3n−1−1⋂k=0

([0,

3k + 13n

]∪[

3k + 23n ,1





C = [0,1] \∞⋃

n=1

3n−1−1⋃k=0

(3k + 1

3n ,3k + 2

3n

)



C =∞⋂

n=1

3n−1−1⋂k=0

([0,

3k + 13n

]∪[

3k + 23n ,1





C = [0,1] \∞⋃

n=1

3n−1−1⋃k=0

(3k + 1

3n ,3k + 2

3n

)



C =∞⋂

n=1

3n−1−1⋂k=0

([0,

3k + 13n

]∪[

3k + 23n ,1





C = [0,1] \∞⋃

n=1

3n−1−1⋃k=0

(3k + 1

3n ,3k + 2

3n

)


Properties and Proofs

Now we will prove some interesting properties of C.


Property 1

Let x = 0.a1a2a3... be the base 3 expansion of a numberx ∈ [0,1]. Then x ∈ C iff an ∈ {0,2} for all n ∈ N

Before we prove property 1, we first have a look at base 3expansion of a number.


Property 1

Let x = 0.a1a2a3... be the base 3 expansion of a numberx ∈ [0,1]. Then x ∈ C iff an ∈ {0,2} for all n ∈ N

Before we prove property 1, we first have a look at base 3expansion of a number.


Fractions in base b

A fraction N in base b is represented in terms of the negativepowers of b:N = a1b−1 + a2b−2 + a3b−3 + ...+ anb−n (∀i ∈ [0,n],ai ∈ (0,b))

How do we convert 0.a1a2a3...10 to another base?

(1) First, pick up the coefficients a1,a2, ...an.(2) Multiply 0.a1a2a3...10 by b.(3) The integer part of the result is a1.(4) If the remaining part is zero, stop.(5) Otherwise, let m = 2.(6) Multiply the remaining part by b.(7) The integer part of the result is am. If the remaining part iszero, stop.(8) Otherwise, let m = m + 1 and goto step (6).


Fractions in base b





Fractions in base b





Example:

Convert 0.37510 to base 2

2N = 2× 0.375 = 0.75 −→ a1 = 0, r1 = 0.75− b0.75c = 0.75

2r1 = 2× 0.75 = 1.5 −→ a2 = 1, r2 = 1.5− b1.5c = 0.5

2r2 = 2× 0.5 = 1 −→ a3 = 1, r3 = 1− b1c = 0

Since r3 = 0, we stop. 0.37510 = 0.0112


Proof for property 1

Let x = 0.a1a2a3... be the base 3 expansion of a numberx ∈ [0,1].

From the way to convert a fraction to another base, we see thateach an corresponds to which third the number is in.

For example, for the number 0.2013, 2 means that it is in the

third third of [0,1], 0 means that it is in the first third of[

23,1]

(the third third of [0,1]), and 1 means that it is in the second

third of[

23,59

](the first third of

[23,1]).







23,1]


third of[

23,59


[23,1]).







23,1]


third of[

23,59


[23,1]).







23,1]


third of[

23,59


[23,1]).


Assume that there exists some k ∈ N such that ak = 1, then xwill be in the middle third of some interval whose middle thirdwill be removed, which means that x /∈ C

On the other hand, by the definition of base 3 expansion, ifan ∈ {0,2} for all n ∈ N, x will never be in the middle third ofany interval whose middle third will be removed. Thus, x ∈ C.


Assume that there exists some k ∈ N such that ak = 1, then xwill be in the middle third of some interval whose middle thirdwill be removed, which means that x /∈ C

On the other hand, by the definition of base 3 expansion, ifan ∈ {0,2} for all n ∈ N, x will never be in the middle third ofany interval whose middle third will be removed. Thus, x ∈ C.


Property 2

The Cantor set is uncountable.

To prove property 2, we need to use the concept of surjectivefunction and cardinality.

Before we prove property 2, we first have a look at surjectivefunction and cardinality.


Property 2





Property 2





Surjective Function and Cardinality

Definition: a function f with domain X and codomain Y issurjective if for every y in Y there exists at least one x in X suchthat f (x) = y .

The cardinality of the domain of a surjective function is greaterthan or equal to the cardinality of its codomain, that is, iff : X → Y is a surjective function, then X has at least as manyelements as Y.


Surjective Function and Cardinality

Definition: a function f with domain X and codomain Y issurjective if for every y in Y there exists at least one x in X suchthat f (x) = y .

The cardinality of the domain of a surjective function is greaterthan or equal to the cardinality of its codomain, that is, iff : X → Y is a surjective function, then X has at least as manyelements as Y.



To show that the Cantor set is uncountable, we need toconstruct a function f from the Cantor set C to the closedinterval [0,1] that is surjective.

Consider the point in C in terms of base 3.

From property 1, we have that for any x = 0.a1a2...3 ∈ [0,1],x ∈ C iff an ∈ {0,2} for all n ∈ N

















Then we construct a function f : C → [0,1] which replaces allthe 2s by 1s, and interprets the sequence as a binaryrepresentation of a real number. In a formula,

f

( ∞∑k=1

ak3−k

)=∞∑

k=1

ak2−k

2

.

For any number y in [0,1], its binary representation can betranslated into a ternary representation of a number x in C byreplacing all the 1s by 2s, so the range of f is [0,1]. Thus, thecardinality of C is greater than or equal to the cardinality of[0,1], which means that C is uncountable.


Then we construct a function f : C → [0,1] which replaces allthe 2s by 1s, and interprets the sequence as a binaryrepresentation of a real number. In a formula,

f

( ∞∑k=1

ak3−k

)=∞∑

k=1

ak2−k

2

.

For any number y in [0,1], its binary representation can betranslated into a ternary representation of a number x in C byreplacing all the 1s by 2s, so the range of f is [0,1]. Thus, thecardinality of C is greater than or equal to the cardinality of[0,1], which means that C is uncountable.


Property 3

The Cantor set has a length of zero, which means that it has nointervals.



We will prove C has a length of zero by showing that the lengthof the complement of C relative to [0,1] is 1.

From the construction of C, we see that at the nth step, we are

removing 2n−1 intervals, all of which are of length13n .

The sum of the length of all intervals removed is

∞∑n=1

2n−1(

13n

)=

13

∞∑n=0

23

n=

13

1

1− 23

= 1

Thus, the length of the complement of C relative to [0,1] is 1,which means that C has a length of zero.







∞∑n=1

2n−1(

13n

)=

13

∞∑n=0

23

n=

13

1

1− 23

= 1








∞∑n=1

2n−1(

13n

)=

13

∞∑n=0

23

n=

13

1

1− 23

= 1








∞∑n=1

2n−1(

13n

)=

13

∞∑n=0

23

n=

13

1

1− 23

= 1








∞∑n=1

2n−1(

13n

)=

13

∞∑n=0

23

n=

13

1

1− 23

= 1



Thank you

Thank you all for listening to my presentation.

Questions?

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Cantor Set and Its Properties - UC Santa Barbaraweb.math.ucsb.edu/~padraic/ucsb_2013_14/mathcs103_s2014/mathcs103... · TitleAbstractPreliminariesConstruction and FormulaProperties